EXPLANATION OF ARSENIC IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) May 29, 2015 Arsenic is a chemical element with symbol As and atomic number 33. Unlike for hydrogen of one electron, a closed-form solution to the Schrödinger equation for the many-electron atoms like the Arsenic atom has not been found. So, various approximations, such as the Hartree–Fock method, could be used to estimate the ground state energies. Under these difficulties I published my paper “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures" (2008) by nalyzing carefully the electromagnetic interactions of two spinning electrons of opposite spin which give a simple formula for the solution of such ground state energies. Under this condition for the electron configuration of Arsenic atom we may write this picture, because in the electron structure of arsenic we observe the following actual electron configuration: 1s22s22p63s23p63d104s24p3 . According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of arsenic (from (E1 to E5) are the following: E1 = 9.8, E2 = 18.6, E3 = 28.35. E4 = 50.13 and E5 = 62.63 . Firstly we examine the - ( E1 + E2 + E3 ) = - 56.75 = E(4p3) Here the E(4p3) represents the binding energy of the three outermost electrons (4p3). Then we observe that -( E4 + E5 ) = -112.76 = E(4s2) It is of interest to note that in the absence of data (from E6 to E33 ) one can write the following theoretical ionizations related to the ground state energies: -( Ε6 +…+ E15 ) = E(3d10) - ( E16 +…+ E21 ) = E(3p6). -(E22 + E23 ) = E(3s2) -(E24 +…+ E29 ) = E(2p6) -(E30 + E31) = E(2s2) -(E32 + E33) = E(1s2) Such theoretical ionization energies are analogous to the experimental values of the ionizations of copper. ( See my EXPLANATION OF COPPER IONIZATIONS ). For understanding better the ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” published in Ind. J. Th. Phys. (2008). ' ' EXPLANATION OF - (E1 + E2 + E3 ) = - 56.75 = E(4p3) Here the binding energy, E(4p3), of the three outermost electrons, (4p3), of parallel spin is given by applying the Bohr formula . The charges (-30e) of the electrons (1s22s22p63s23p63d104s2 ) screen the nuclear charge (+33e) and for a perfect screening we would have an effective ζ = 3. However the three electrons (4p3 ) repel the electrons of 4s2 leading to the deformation of spherical electron clouds. Thus ζ > 3. Under this condition we may write ( E1 + E2 + E3 ) = 56.75 eV = -E(4p3) = - 3(- 13.6057 )ζ2 / n2 Since n = 4 we get ζ = 4.71 > 3 . EXPLANATION OF -( Ε4 + E5 ) = - 112.76 = E(4s2) ''' Here the E(4s2) represents the binding energy of the 2 electrons with opposite spin given by applying my formula of 2008. '''The charges (-28e) of the inner electrons (1s22s22p63s23p63d10) screen the nuclear charge (+33e) and for a perfect screening we would have ζ = 5. However the electrons of 4s2 penetrate the 3d10 and lead to the deformations of spherical electron clouds. Thus we must observe that ζ > 5. Under this condition we may write ( Ε4 + E5 ) = 112.76 eV = -E(4s2) = - + (16.95)ζ - 4.1 / n2 Since n = 4 the above equation could be written as 1.7ζ2 - 1.06ζ - 112.5 = 0 Then solving for ζ we get ζ = 8.45 > 5 . Note that the two electrons of opposite spin (4s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. However in the absence of adetailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. Category:Fundamental physics concepts